https://artofproblemsolving.com/wiki/index.php?title=2005_IMO_Shortlist_Problems/G3&feed=atom&action=history2005 IMO Shortlist Problems/G3 - Revision history2024-03-29T10:16:15ZRevision history for this page on the wikiMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2005_IMO_Shortlist_Problems/G3&diff=15243&oldid=prevBoy Soprano II at 14:53, 24 June 20072007-06-24T14:53:19Z<p></p>
<p><b>New page</b></p><div>== Problem ==<br />
<br />
(''Ukraine'')<br />
Let <math> \displaystyle ABCD </math> be a parallelogram. A variable line <math> \ell </math> passing through the point <math> \displaystyle A </math> intersects the rays <math> \displaystyle BC </math> and <math> \displaystyle DC </math> at points <math> \displaystyle X </math> and <math> \displaystyle Y </math>, respectively. Let <math> \displaystyle K </math> and <math> \displaystyle L </math> be the centres of the excircles of triangles <math> \displaystyle ABX </math> and <math> \displaystyle ADY </math>, touching the sides <math> \displaystyle BX </math> and <math> \displaystyle DY </math>, respectively. Prove that the size of angle <math> \displaystyle KCL </math> does not depend on the choice of <math> \ell </math>.<br />
<br />
''This was also Problem 3 of the 2006 3rd German TST, and a problem at the 2006 India IMO Training Camp. It also appeared in modified form as Problem 3, Day 3 of the 2006 Iran TST.''<br />
<br />
== Solution ==<br />
<br />
Let <math> \ell_1, \ell_2 </math> be the interior angle bisectors of <math> \displaystyle ABX, YAD </math>. Let <math> \displaystyle m_1, m_2 </math> be the exterior angle bisectors of <math> \displaystyle ABC, CDA </math>. Then <math> \displaystyle K </math> is the intersection of <math> \ell_1, m_1 </math> and <math> \displaystyle L </math> is the intersection of <math> \ell_2, m_2 </math>.<br />
<br />
<center> [[Image:ISL2005G3.png]] </center><br />
<br />
Let us denote <math> \displaystyle x,y </math> as the measures of <math> \angle BAK, \angle LAD </math>, and denote <math> \displaystyle \alpha = x+y </math>. Then <math> \angle BAD \equiv \angle DCB \equiv 2 \alpha </math>. Furthermore, since <math> \displaystyle BK </math> is the exterior angle bisector of <math> \displaystyle ABC </math>, we know that the exterior angle at <math> \displaystyle ABK </math> is <math> \displaystyle \alpha = x+y </math>, so <math> \angle AKB \equiv y </math>. Similarly, <math> \angle ALD \equiv x </math>. It follows that triangles <math> \displaystyle ABK, LDA </math> are similar. Then since <math> \displaystyle ABCD </math> is a parallelogram,<br />
<center><br />
<math> \frac{BK}{DC} = \frac{BK}{AB} = \frac{DA}{LD} = \frac{CB}{LD} </math>.<br />
</center><br />
Since we know <math> \angle KBC \equiv \angle CDL \equiv \alpha </math>, this implies that triangles <math> \displaystyle KBC, CDL </math> are similar. This means that<br />
<center> <br><br />
<math> \angle KCL \equiv 2\pi - (\angle LCD + \angle DCB + \angle BCK) \equiv 2\pi - [(\angle CBK + \angle BCK) + \angle DCB] </math> <math> \equiv 2\pi - [ (\pi- \alpha) + 2\alpha] \equiv \pi - \alpha </math>,<br />
</center> <br><br />
which is independent of <math> \ell </math>, since <math> \displaystyle \alpha </math> is always half the measure of <math> \angle BAD </math>. {{Halmos}}<br />
<br />
<br />
{{alternate solutions}}<br />
<br />
== Resources ==<br />
<br />
* [[2005 IMO Shortlist Problems]]<br />
* [[2006 Germany TST Problems]]<br />
* [[2006 Iran TST Problems]]<br />
* [[2006 IMOTC Problems]]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=512720#512720 Discussion on AoPS/MathLinks]<br />
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=498848#498848 More Discussion]<br />
<br />
<br />
[[Category:Olympiad Geometry Problems]]</div>Boy Soprano II